Quantum Mechanics MCQ Quiz - Objective Question with Answer for Quantum Mechanics - Download Free PDF

Calculation:

In case of spin-1/2 fermions:

E_F = 2E₁ + E₂ = 2E₁ + 4E₁ = 6E₁

In case of Bosons:

E_B = 3E₁

Thus, the ratio is:

E_F / E_B = 6E₁ / 3E₁ = 2

Calculation:

The commutation relation [x, p] = iℏ cannot be satisfied if the dimension of the Hilbert space is finite. In finite Hilbert space, x and p can be written as finite matrix terms. But for any operator A and B having finite matrix elements:

trace([A, B]) = 0

trace(A B) = trace(B A)   Since [A, B] = A B - B A

Therefore, it can't be constant.

Calculation:

L × L = (L_x + L_y + L_z) × (L_x + L_y + L_z)

Expanding the terms:

= (L_yL_z - L_zL_y) + (L_zL_x - L_xL_z) + (L_xL_y - L_yL_x)

= [L_y, L_z] + [L_z, L_x] + [L_x, L_y]

Using commutation relations:

[L_y, L_z] = iℏL_x, [L_z, L_x] = iℏL_y, [L_x, L_y] = iℏL_z

Thus, L × L = iℏL

Explanation:

The particle in bound state is moving back and forth, and so its average momentum for any quantum state is zero if ψ is real.

Calculation:

ψ(x) = 2√3 e^-x(1 - e^-x) ; x ≥ 0

ψ(x) = 0 ; x

⇒ |ψ|² = [2√3 (e^-x - e^-2x)]² = 4 × 3 (e^-2x + e^-4x - 2e^-3x)

⇒ |ψ|² = 12 (e^-4x + e^-2x - 2e^-3x)

⇒∫₀^∞ |ψ|² dx = -12 [(-1/4) - 1/2 + 2/3] = -12 [(-3 - 6 + 8) / 12] = 1

For most probable position: d(|ψ|²)/dx = 0

⇒ -4e^-4x - 2e^-2x + 6e^-3x = 0

⇒ 4e^-2x - 6e^-x + 2 = 0       (Let e^-x = y)

⇒ 4y² - 6y + 2 = 0

⇒ 2y(y - 1) - 1(y - 1) = 0

⇒ (y - 1)(2y - 1) = 0

⇒ y = 1 and y = 1/2

⇒ e^-x = 1 and e^-x = 1/2

x_p = 0.693

⟨x⟩ = ∫ x|ψ|² dx = 12 [ ∫₀^∞ x e^-4x dx + ∫₀^∞ x e^-2x dx - 2∫₀^∞ x e^-3x dx ]

⇒ ⟨x⟩ = 12 (1/16 + 1/4 - 2/9) = 12 [(9 + 36 - 32) / 144] = 13/12 = 1.08

x_p = 0.693

⟨x⟩ / x_p = 1.08 / 0.693 = 1.563

Concept:

We are using properties of angular momentum viz-=^2+^2+^2\) and by putting given values of  and  we can find value of .

Explanation:

 Given,

•  =6\hbar^2|\phi>\)
• =2\hbar|\phi>\)

Here L is angular momentum and  is Planck's constant

Using angular momentum formula, we can write expectation values of angular momentum as

• =^2+^2+^2\)

Applying ket, bra operator on angular momentum operator, we get

• =++\)

• =++\)

Using, =1\), and putting given values of  and we get,

• ++4\hbar^2\)
• Now, \)\)

we get, \)

• =\hbar^2\)

The correct answer is =\hbar^2\)

Concept:

The given Hamiltonian is for the Anharmonic Oscillator. We will compare the given Hamiltonian with the equation of the Hamiltonian Anharmonic oscillator.

Explanation:

• , 

This is the formula for energy in an oscillator in two-dimension.

Now, (given)

• Substitute values of  and  in terms of 
•  should be even to satisfy the equation.
•  ; 
• So, the degeneracy of the energy level  is 7.

So, the correct answer is 7.

Explanation:

• The wave function for each particle can be written in terms of the energy eigenstates of the infinite square well. The ground state and the first excited state of a particle (in a one-dimensional infinite square well of width $a$) are given by:

 , for the ground state, and

 , for the first excited state.

• The expectation value of the product of the position operators  and  can be calculated with the help of the integral:

 ,

• Where the integration is over the range [0, a] for both  and .
• However, since the particles are distinguishable and non-interacting, we can write , which simplifies the integrations to:

 , and .

• Calculation of these integrals gives: , and .
• So, the expectation value of the product of  and  is just the product of the expectation values of  and : 

Solution-Option-1

Concept:

First, we will check the number of radial nodes in the graph which is given by

• Number of radial nodes 
• Then secondly we will check the value of probability of electron at .

Calculation-

• Given-  and 
• Radial nodes 
• So, there will be only one node in the graph. Either option 1 is correct or option 2 is correct.
• Now we will check the probability of finding an electron at .
• At,    
• At , 
• The probability of finding an electron at ">r=2a" id="MathJax-Element-14-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a" id="MathJax-Element-270-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a" id="MathJax-Element-124-Frame" role="presentation" style="position: relative;" tabindex="0">r=2ar=2a $r=2a$  is zero.
• So, graph 1 satisfies this condition.

So, the correct answer is Graph-1.

Explanation:

 (given)

Case-1-Initial ground state energy without polarization

According to Pauli Exclusion principle, only two electrons filled in one state.

• Initial ground state energy 
• Now, 

Case-2-Final ground state energy after polarization

After polarization, only one electron filled in the state.

The change in ground state energy is 

So, the correct answer is .

Concept:

The momentum operator is given by

p = - ih 

where h is the Plank constant.

Calculation:

e |x>

= [ ]|x>

= |x>

= |x> - a∇|x> +  (a∇)²|x> ... = |x-a>

X|x-a> = (x-a)|x-a>

The correct answer is an option (3).

CONCEPT: 

1. Energy Operator:

The energy (or Hamiltonian) operator in one dimension for a particle in a box is given by:

2. Expectation Value of Energy:

The expectation value of energy ⟨ E ⟩  is given by:

⟨ E ⟩ =

Calculation -

1. Wavefunction:
   

2. Second Derivative:
   

   

Simplifying:

3. Hamiltonian Acting on ψ(x):
   
   

4. Expectation Value Integral:
   
⟨ E ⟩ =

The normalized wavefunction ψ(x) given is a linear combination of eigenfunctions of the infinite potential well.

For an infinite potential well, the eigenfunctions are

with energy eigenvalues

Comparing the given wavefunction:



This is equivalent to a superposition of the first and second eigenstates.

The coefficients and normalization ensure that this wavefunction is a proper eigenstate mixture.

Energy Calculation:

By symmetry and orthogonality of the eigenfunctions, the expectation value of energy ⟨ E ⟩ is the weighted sum of eigenvalues:

⟨ E ⟩ =

Given:

The weights a₁ , a₂ are found from normalization. Simplifying using known integrals, we obtain the correct weighted sum.

Finally, the result for this particular problem (via solving) yields the expectation value 

Therefore, the correct answer is (2).

Explanation:

Let us reconsider the system of equations:

 and 

By differentiating the first equation again, 

• Substituting the second equation into this results in 
• This differential equation is a simple harmonic one, but with a key difference: there is no negative in front of ω², leading to hyperbolic solutions.
• Specifically, we find A(t) = Csinh(ωt), for some constant C.
• Given that the expectation value  at t = 0, we find 
• Thus, in general, B(t) has to be in the form of cosh(ωt), to meet the commutation relations. Finally, given that  we need to multiply cosh(ωt) by i.
• So the time-evolved expectation value is 

Concept: 

 Energy in first perturbation is given by

• \)
• We know that, for an infinite potential well,  

Explanation:

Given-

•  limits from  to 

We know that, for an infinite potential well,  

• \)

Now,  and , put these values in first order energy perturbation equation, we get,

Now for changing the limit from( to ) to ( to )

Using the trignometric formula for ,

• Substituting this value, we get,

So, the correct answer is .

Concept:

A generator is an operator that acts on the wave function or quantum state vector, to produce the effect of applying a small transformation to the system.

Calculation:

q → q' = (1 + ϵ)q

p → p' = (1 - ϵ)p

If G is the generator then 

p' - p = δ p_j 

= - ϵ 

= - ϵ p

q' - q = δ q_i 

= ϵ 

= ϵ p

Now G = qp

- ϵ  = - ϵ p = δ p

ϵ  = - ϵ q = δ q

The correct answer is option (2).